# Calculate the sum of the solutions and the sum of the squares of the solutions. Given x1,x2,x3,x4 the solutions of the equation x^4+1=0.

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### 2 Answers

You also may solve the equation in complex set of numbers, such that:

`x^4 + 1 = x^2 - (-1)`

Since using complex number theory yields `-1 = i^2` , then you may substitute `i^2` for 1 in equation, such that:

`x^4 - i^2 = 0 `

Converting the difference of squares into a product, yields:

`(x^2 - i)(x^2 + i) = 0 => {(x^2 - i = 0),(x^2 + i = 0):}`

`{(x^2 = i),(x^2 = -i):} => {(x^2 = +-sqrt i),(x^2 = i^3):} => {(x^2 = +-sqrt i),(x^2 = +-i*sqrt i):}`

Hence, evaluating the sum of solutions, yields:

`x_1 + x_2 + x_3 + x_4 = sqrt i - sqrt i + isqrt i - i sqrt i`

`x_1 + x_2 + x_3 + x_4 = 0`

`(x_1 + x_2 + x_3 + x_4)^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + 2(x_1*x_2 + x_1*x_3 + x_1*x_4 + x_2*x_3 + x_2*x_4 + x_3*x_4)`

`x_1^2 + x_2^2 + x_3^2 + x_4^2 = (x_1 + x_2 + x_3 + x_4)^2 - 2(x_1*x_2 + x_1*x_3 + x_1*x_4 + x_2*x_3 + x_2*x_4 + x_3*x_4)`

`x_1^2 + x_2^2 + x_3^2 + x_4^2 = 0^2 - 2*(-i - 1 + 1 + 1 - 1 + i)`

`x_1^2 + x_2^2 + x_3^2 + x_4^2 = 0 - 2*0`

`x_1^2 + x_2^2 + x_3^2 + x_4^2 = 0`

**Hence, evaluating the indicated summations yields `x_1 + x_2 + x_3 + x_4 = 0` and `x_1^2 + x_2^2 + x_3^2 + x_4^2 ` `= 0` **.

We'll apply Viete expressions, which are the connection between roots and coefficients of an equation:

x1 + x2 + x3 + x4=0

x^2 + x2^2 + x3^2 + x4^2 = (x1 + x2 + x3 + x4)^2 - 2(x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4)

x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4= c/a=0/1=0

x^2 + x2^2 + x3^2 + x4^2 = 0 - 2*0=0